JEE Mains and Advanced

Numerical 1:

• Let (p) be the frequency of the dominant gene.
• (p^2) is the frequency of affected individuals.
• So, (p^2 = 0.01 \Rightarrow p = 0.1)
• Frequency of carriers (= 2pq = 2 \times 0.1 \times 0.9 = 0.18 ) or (18%).

Numerical 2:

• Let (μ = 10) and (σ = 2).
• Sample size (= n = 100).
• Probability of sample mean between 9.5 mm and 10.5 mm (= P(9.5 < \overline{X} < 10.5)).
• Using CLT, (P \left (\frac{9.5-10}{0.2} < \frac{\overline{X} - \mu}{σ/\sqrt{n}} \frac{10.5-10}{0.2} \right )).
• (P(-2.5 < Z < 2.5) = 0.99) (using standard normal tables).

CBSE Board Exams

Numerical 1:

• Tall allele: (T), Short allele: (t), Dominant Tall: (TT), Recessive Short: (tt), Heterozygous: (Tt).
• In Hardy-Weinberg equilibrium:
  • (p^2) is the frequency of homozygous dominant type,
• (q^2) is the frequency of recessive homozygous type,
• (2pq) is the frequency of heterozygous type, and (p+q=1).
• Given that 75% are tall, so frequency of dominant allele (p^2 = 0.75), so (p=0.87) and (q=1-p=0.13).
• Expected tall plants: (p^2 + 2pq = (0.87)^2 + 2(0.87)(0.13) = 0.76 + 0.224 = 0.984), which is roughly (76).
• Expected short plants: (q^2 = (0.13)^2 = 0.0169) or roughly (2).

Numerical 2:

• Similar to Numerical 1, we get the following;
• (p^2 =0.75 \Rightarrow p=0.87)
• (q^2 = 0.25 \Rightarrow q=0.5).
• So expected number of butterflies with brown wings (= 0.87^2 + 0.87 \times 0.5 \times 2= 0.76 + 0.87 = 0.933) i.e., roughly (93).
• Similarly, expected number of butterflies with white wings (= 0.5^2 = 0.25 ) i.e., roughly (25).